The mass of the object would be 52.5g
Answer:
C. 178−−−√ m
Step-by-step explanation:
Given the following :
v = final velocity (in m/s)
u = initial velocity (in m/s)
a = acceleration (in m/s²)
s = distance (in meters).
Find v when u is 8 m/s, a is 3 m/s², and s is 19 meters
Using the 3rd equation of motion :
v^2 = u^2 + 2as
v^2 = 8^2 + 2(3)(19)
v^2 = 64 + 114
v^2 = 178
Take the square root of both sides :
√v^2 = √178
v = √178
Answer:
2apply.
Step-by-step explanation:
Answer: The flagpole is about 22.3 feet tall.
This is roughly the size of a two story building, assuming each floor is 10 feet or so.
Since this height is under 25 ft, this flagpole is in compliance with the regulation.
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Work Shown:
The horizontal leg of the triangle is 36 ft. This is the adjacent leg to the reference angle 25 degrees.
The vertical leg of the triangle is x-5.5, and this leg is the opposite side to the reference angle 25 degrees.
Use the tangent rule to connect the opposite and adjacent sides.
Solve for x.
tan(angle) = opposite/adjacent
tan(25) = (x-5.5)/36
36*tan(25) = x-5.5
x-5.5 = 36*tan(25)
x = 36*tan(25)+5.5
x = 22.28707569358
x = 22.3 ft is the approximate height of the flagpole
Make sure your calculator is in degree mode.
The height is not over 25 ft, so this flagpole meets the requirements.
Answer:
The answer is k = 4.
Step-by-step explanation:
Given:
Y = 8 and X = 2.
Now, to solve direct variation.
direct with 
.
Now, using equation to solve direct variation:
Dividing both sides by 2 we get:
.
Therefore, the answer is k = 4.
